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# Can You?

I was wondering if you can use a 8 ohm and a 4 ohm woofer in the same cabinet.

Is there any danger to your amp?

Does it cause some sort of distortion?

OR anything else.

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You would have more output from the 4ohm driver all other things being equal, so it would be doing more of the work.
Danger to the amp, well that would depend on the wiring, but normally no. With a 4 & 8 ohm combo you would see is 6 ohms if wired in parallel, or 10ohm if in series.
Either way I would recommend against it.
It can be done, assuming you are designing the crossover properly. Often, the impedance of tweeters aren't the same as the midwoofers they are designed to work with. But then, if you were building a crossover you would have already known that...

Quote:
Originally Posted by sonytheater

With a 4 & 8 ohm combo you would see is 6 ohms if wired in parallel, or 10ohm if in series.

Series is the easiest calcualtion. Simple addition 8 ohms +4 ohms = 12 ohms. Parallel is (1/8) + (1/4) = 2.66 ohms.

-Robert
Quote:
Originally Posted by rlj5242

Series is the easiest calcualtion. Simple addition 8 ohms +4 ohms = 12 ohms. Parallel is (1/8) + (1/4) = 2.66 ohms.

-Robert

Da, ya got me for the series not sure where I was on that. But I dont understand where your getting the 2.66? I've wired quite a few odd configurations & cant recall ever getting something like that from a 8/4 combo.
Quote:
Originally Posted by sonytheater

Where did you get that formula? I've never seen anything come out like that & I've wired quite a few odd configurations.

Ummm...it's called Ohm's law. You know, V=IR?

If you have a constant voltage across two parallel resistances, the currents add together for the total current, right? When you add the two currents and back-calculate for the combined effective load, it becomes immediately clear that the effective combined parallel resistance must be lower than either of the two individual resistances. QED....do the math.
Quote:
Originally Posted by MauneyM

Ummm...it's called Ohm's law. You know, V=IR?

If you have a constant voltage across two parallel resistances, the currents add together for the total current, right? When you add the two currents and back-calculate for the combined effective load, it becomes immediately clear that the effective combined parallel resistance must be lower than either of the two individual resistances. QED....do the math.

Hay theres no reason to be a smart ass, so if you cant restrain your self to a simple answer than just don't answer.
rlj5242, is correct.
Quote:

Hay theres no reason to be a smart ass,

he wasn't
Well thats not how it looked to me.
Quote:
Originally Posted by sonytheater

Well thats not how it looked to me.

See above - emoticon added for clarity.
Quote:
Originally Posted by sonytheater

Da, ya got me for the series not sure where I was on that. But I dont understand where your getting the 2.66? I've wired quite a few odd configurations & cant recall ever getting something like that from a 8/4 combo.

There's a missing reciprocal here -

For resistors in parallel, R = 1/(1 / r1 + 1/r2 ... + 1/rn)

1/(1/4 + 1/8) = 8/3 = 2.6 repeating.

a simplifying assumption : that the drivers are purely resistive loads, which they're not. Driver impedance has reactive components with frequency dependant magnitutde. The calculation only gives us a possible minimum impedance.

and an assumption that may be wrong: the drivers do not have filters in front of them. Capacitor impedance is inversely proportional to frequency while inductor impedance is proportional to frequency. So in a 2-way speaker the impedance of each driver plus its cross-over filter (hi-pass or low-pass) is increasing as the other's is decreasing, and the minimum impedance may never drop below the worst of the two drivers.
Quote:
Originally Posted by Drew Eckhardt

There's a missing reciprocal here -

For resistors in parallel, R = 1/(1 / r1 + 1/r2 ... + 1/rn)

1/(1/4 + 1/8) = 8/3 = 2.6 repeating.

a simplifying assumption : that the drivers are purely resistive loads, which they're not. Driver impedance has reactive components with frequency dependant magnitutde. The calculation only gives us a possible minimum impedance.

and an assumption that may be wrong: the drivers do not have filters in front of them. Capacitor impedance is inversely proportional to frequency while inductor impedance is proportional to frequency. So in a 2-way speaker the impedance of each driver plus its cross-over filter (hi-pass or low-pass) is increasing as the other's is decreasing.

Ok well Im lost in all the technical stuff here but Ill take your word on it
Call me a moron or what you will for it but Im a hands on & not a by the book guy.
Just as an example I had a few people tell me I couldent get the right impedance with my 6 driver IB. Everyone was saying it would fall either somewhere at or under 2ohm, or above 8ohm. Well after playing around a bit I got to a 3.72ohm load, with all 12 VC's wired.
Here's what you should use - a calculator free and available online.
http://www.the12volt.com/caraudio/boxcalcs.asp

I use these calcs, along with others, all the time.

And I agree, he was a smartass with that answer.
Quote:
Originally Posted by stevdart

And I agree, he was a smartass with that answer.

How? I gave a very detailed, explicit explanation of why the first answer was correct (even if his described equation was wrong). Are we supposed to say that wrong is right in order to avoid being labelled a smartass??
That's okay, I'm a smartass half the time myself. It's not necessarily a negative thing, but it is what it is. In fact, I relish in my smartass abilities!

Your response was really just borderline smartass. I ammend my remark to include the descriptive "borderline"; insert as appropriate.

It was the "QED...do the math" part, by the way, that pushed it into smartass territory. I figure if I can't understand quickly what the letters stand for, the OP probably can't either. And "do the math" is always smartass, whenever it is said. True that?
Quote:
Originally Posted by stevdart

It was the "QED...do the math" part, by the way, that pushed it into smartass territory. I figure if I can't understand quickly what the letters stand for, the OP probably can't either. And "do the math" is always smartass, whenever it is said. True that?

Ahhhhh....Now I see the issue.

QED - Quod Erat Demonstrandum - "Which was to be shown". This is an old latin math/logic term used at the end of a proof or a description, basically saying "that's all you need to prove the point", or "proven as requested". Thus "QED - do the math" means "that's enough info to go get the specifics and work the problem in real life".

Never intended to be smartass.....
Quote:

Well after playing around a bit I got to a 3.72ohm load, with all 12 VC's wired.

How do you know? How did you measure the combined impedance?
Don't want to stir things up, but just to clarify for thost that are having trouble understanding...

A good analogy for resistance and impedance is to think of it as trying to shove water through a short and narrow pipe. The smaller the pipe, the more resistance (ohms).

If you want to connect 2 (identical) pipes together, you have two choices...

* Series, where you connect the second one after the first, so that all water must go through both narrow pipes. This doubles your resistance/impedance, as all water must go through two narrow pipes.

* Parallel, where you connect them side-by-side, allowing some water to go through one, and the other other water through the other one. This is easier, because each pipe has only half as much water going through it now (compared to if there were only one pipe). This has the effect of cutting the resistance in half.

If the pipes are not identical, you can calculate as described above.
Series, you just add them together
Parallel, you invert them, add them, invert the total. 1 / (1/R1 + 1/R2). Also as mentioned above, you will always get a number less than your smallest input.
Quote:
Originally Posted by Drew Eckhardt

a simplifying assumption : that the drivers are purely resistive loads, which they're not. Driver impedance has reactive components with frequency dependant magnitutde. The calculation only gives us a possible minimum impedance.

and an assumption that may be wrong: the drivers do not have filters in front of them. Capacitor impedance is inversely proportional to frequency while inductor impedance is proportional to frequency. So in a 2-way speaker the impedance of each driver plus its cross-over filter (hi-pass or low-pass) is increasing as the other's is decreasing, and the minimum impedance may never drop below the worst of the two drivers.

This is an important point - the impedance of the drivers will be different at different frequencies. It's hard to give any specific advice without more specific input from the OP. Are these woofers being used for a speaker or for a subwoofer? Do you have more specific information on the woofers?

Generally speaking, though, I think this advice was appropriate:
Quote:
Originally Posted by stevdart

It can be done, assuming you are designing the crossover properly. Often, the impedance of tweeters aren't the same as the midwoofers they are designed to work with. But then, if you were building a crossover you would have already known that...

I replaced some speakers in some old cabinets.

It originally had two 8inch drivers at 4ohms a piece and 4 2 1/2inch (each 4 ohms) tweeters when connected it yielded 8 total ohms with the crossover. I replaced one woofer in each with a full range pioneer speaker (Here) because the higer end of the midrange was terrible for these speakers.

Now with the 8ohm pioneer the total resistance comes out to around 11ohms. When I place a meter on the terminals it fluctuates between 10 and 12 ohms usually coming around to about 11.

So what do you think? My kenwood reciever (vr-309) can handle 100w (8ohm -16ohm) and the speakers now sound decent with the current configuration. However, I would like to get them back to 8 ohms.

Is there a way that I could hook up a resistor in parallel with the pioneer to make it a 4 ohm connection with the other woofer. if so would there be any performance loss or consequences

Thanks for all the info so far.
Quote:
Originally Posted by msheabel

My kenwood reciever (vr-309) can handle 100w (8ohm -16ohm) and the speakers now sound decent with the current configuration.

Given that you've changed the load impedance your crossover is seeing, if they sound 'decent', you've been lucky.

Quote:

However, I would like to get them back to 8 ohms.

Why? You amp may output a hair more power, but it's not likely to make a huge difference.

Quote:

Is there a way that I could hook up a resistor in parallel with the pioneer to make it a 4 ohm connection with the other woofer. if so would there be any performance loss or consequences

Yes, you could put an 8 Ohm load resistor in parallel with the new 8 Ohm speaker to create a 4 Ohm load (just like the original driver you took out). However, this will probably reduce the output of the speaker somewhat (by half - 3 db), and will definitely affect the balance between the various drivers. Though the crossover will now see the load it originally expected, half of the power that is being sent to this driver will now be dissipated into heat in the resistor, thus reducing your sound output.

Now, the upside is that your crossover will now be seeing the total load impedance it was designed for. Then again, you said you didn't like the sound to begin with, so this may not be an improvement.

Overall, it's an interesting idea, and it might be worth trying to see if it sounds better than what you have. I'd bet it won't (I'd expect reduced low-end output), but there are so many variables we don't know in the crossover and system impedance function.......? Give it a shot and see what happens. Unless you make a wiring error, you won't hurt anything by trying it.

Good luck.....
I agree that you should not worry about getting the reading down to 8 ohms. If it sounds better than the original speaker configuration, stick with it and let it reproduce music for you. In crossover designing, it's usually the case where you have to get the impedance HIGH enough. I wouldn't mind if I end up with 12 or 16 ohms nominal.
Quote:

Now with the 8ohm pioneer the total resistance comes out to around 11ohms.

You're measuring the resistance of the crossover input.

Impedance is what you should be concerned with.
Targus, in your opinion, what's the best way to find the nominal impedance? I have been using the output graph in Speaker Workshop, along with measurement of DCR (at crossover inputs) X 1.3 to form a conclusion. And also noticing if the amplifier clicks into protection with too low of a load
Quote:
Originally Posted by stevdart

Targus, in your opinion, what's the best way to find the nominal impedance? I have been using the output graph in Speaker Workshop, along with measurement of DCR (at crossover inputs) X 1.3 to form a conclusion. And also noticing if the amplifier clicks into protection with too low of a load

I'm not Targus, but wouldn't the best way to be to just measure current and voltage with a steady-state signal (pink noise, perhaps)? V/I?
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